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Câu hỏi: Cho hàm số $y=f\left( x \right)$ thỏa mãn $f\left( 2 \right)=-\dfrac{4}{19}$ và $f'\left( x \right)={{x}^{3}}{{\left[ f\left( x \right) \right]}^{2}},\forall x\in \mathbb{R}.$ Giá trị của $f\left( 1 \right)$ bằng
A. $-\dfrac{2}{3}.$
B. $-\dfrac{1}{2}.$
C. $-1.$
D. $-\dfrac{3}{4}.$
A. $-\dfrac{2}{3}.$
B. $-\dfrac{1}{2}.$
C. $-1.$
D. $-\dfrac{3}{4}.$
Ta có $f'\left( x \right)={{x}^{3}}{{\left[ f\left( x \right) \right]}^{2}}\Leftrightarrow -\dfrac{f'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}=-{{x}^{3}}.$
$\Rightarrow \int\limits_{{}}^{{}}{\left( -\dfrac{f'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}dx \right)=-\int\limits_{{}}^{{}}{{{x}^{3}}dx}\Rightarrow \dfrac{1}{f\left( x \right)}=-\dfrac{{{x}^{4}}}{4}+C.\left( 1 \right)}$
Thay $x=2$ vào (1) ta được $\dfrac{1}{f\left( 2 \right)}=-\dfrac{{{2}^{4}}}{4}+C\Rightarrow C=-\dfrac{3}{4}.$
Vậy $\dfrac{1}{f\left( x \right)}=-\dfrac{{{x}^{4}}}{4}-\dfrac{3}{4}\Rightarrow \dfrac{1}{f\left( 1 \right)}=-\dfrac{{{1}^{4}}}{4}-\dfrac{3}{4}\Rightarrow f\left( 1 \right)=-1.$
$\Rightarrow \int\limits_{{}}^{{}}{\left( -\dfrac{f'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}dx \right)=-\int\limits_{{}}^{{}}{{{x}^{3}}dx}\Rightarrow \dfrac{1}{f\left( x \right)}=-\dfrac{{{x}^{4}}}{4}+C.\left( 1 \right)}$
Thay $x=2$ vào (1) ta được $\dfrac{1}{f\left( 2 \right)}=-\dfrac{{{2}^{4}}}{4}+C\Rightarrow C=-\dfrac{3}{4}.$
Vậy $\dfrac{1}{f\left( x \right)}=-\dfrac{{{x}^{4}}}{4}-\dfrac{3}{4}\Rightarrow \dfrac{1}{f\left( 1 \right)}=-\dfrac{{{1}^{4}}}{4}-\dfrac{3}{4}\Rightarrow f\left( 1 \right)=-1.$
Đáp án C.